Name 3 muscles of the leg

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

If you fire a gun horizontally, and at the same time drop a bullet/slug from the same height, what will happen?

Elena-2011 [213]

Answer:

There you have it. The y-component of air resistance for the fired bullet still depends on the fired speed of the bullet (since it is proportional to v2). A fired bullet (with air resistance) does not hit the ground at the same time as a dropped bullet.

Explanation:

hope this helped ✨

8 0

3 years ago

A motorcyclist is traveling at 58.3 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic

Novosadov [1.4K]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one st

strojnjashka [21]

Answer:

20.45%

Explanation:

The probability that the student got a B is

$\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%$

Now, how many students are there in total?

The answer is

$10+3+16+4+6+5=44\; \text{students}$

How many students got a B?

The answer is

$3+6=9\; \text{students}$

therefore, the probability that the student has got a B is

$\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%$

Hence, the probability that a student has got a B is 20.45%

5 0

1 year ago

A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6

Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

4 0

3 years ago

Name 3 muscles of the leg​

Name 3 muscles of the leg